estatica

6.1 a 6.8 Utilice el método de nodos para determinar la fuerza presente en cada elemento de las armaduras que muestran las fig. siguientes. Establezca si están en tensión o en compresión.



ΣMA =0
Cx (4) – (800 x 7.5) = 0
4Cx – 6000 = 0
Cx = 
Cx = 1500 lb.

ΣFX = 0
Cx – Ax = 0
1500 lb. – Ax = 0
Ax =1500 lb. = FAB (T)

NODO “B”


ΣFX = 0
FCB ( ) – 1500 = 0

FCB = 

FCB = 1700 lb. (C)
ΣFy = 0
FCB ( ) – 800 lb. = 0

1700 lb. ( ) – 800lb. = 0

800 lb. – 800 lb. = 0
0 = 0

NODO “C”


ΣFX = 0
1500 – 1700 lb. ( ) = 0

0 = 0

ΣFy = 0
FAC –FCB ( )= 0

FAC –1700 lb. ( ) = 0

FAC – 800 lb. = 0

FAC = 800 lb. (T)













ΣMA = 0
Cy (6) – (4.2 x 4) = 0
6Cy – 16.8 = 0Cy = 
Cy = 2.8 kN.

ΣFy = 0
2.8 kN – 4.2kN + Ay = 0
Ay – 1.4kN = 0
Ay = 1.4kN

NODO “B”


ΣFX = 0FAB ( ) – FCB ( ) = 0

* FAB ( ) – [7 –  FAB ]( ) = 0

 FAB ( ) – 5.6 +  FAB  = 0

 FAB ( ) – 5.6 = 0
FAB = 
FAB =3.39411255 Kn (C)

ΣFy = 0
FAB ( ) + FCB ( ) – 4.2kN = 0

FCB = 

* FCB = 7 –  FAB 
FCB = 7 –  (3.39411255kN) 
FCB = 7 – 4FCB = 3Kn (C)

NODO “C”


ΣFX = 0
3 ( ) – FAC ( ) = 0

 – FAC ( ) = 0

FAC = 

FAC = 2.60 kN (T)

ΣFy = 0
2.8kN – FCB( ) –FAC() = 0

2.8kN – 3kN( ) – 2.6kN() = 0

2.8kN – 1.8kN – 1kN = 0

0 = 0


ΣMA = 0
Cy (7.5) – (450 x 31.5) = 0
7.5Cy –...